At a certain temperature, 0.820 mol SO3 is placed in a 3.50 L container. 2 SO₂ (g) 2 SO₂(g) + O₂(g) At equilibrium, 0.170 mol O₂ is present. Calculate Kc. Kc =

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At a certain temperature, 0.820 mol SO3 is placed in a 3.50 L container.
2 SO₂(g) 2 SO₂(g) + O₂(g)
At equilibrium, 0.170 mol O₂ is present. Calculate Kc.
Kc =
Transcribed Image Text:At a certain temperature, 0.820 mol SO3 is placed in a 3.50 L container. 2 SO₂(g) 2 SO₂(g) + O₂(g) At equilibrium, 0.170 mol O₂ is present. Calculate Kc. Kc =
Expert Solution
Step 1

Initial conc. of SO3 = moles / volume = 0.820 mol / 3.50 L = 0.23429 M

The equilibrium is

  2 SO3 (g) --------> 2 SO2 (g) + O2 (g)

Since there are no products, the equilibrium will shift to right side (product side)

Let x be the change in the concentration of each species.

                       SO3 (g) -------->   2 SO2 (g) + O2 (g)

Initial              0.23429                    0                0

Change              – 2x                      + 2x           + x

Equilibrium     0.23429 – 2x               2x             x

Equilibrium conc. of O2 = x = 0.170 mol / 3.50 L = 0.04857 M

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